of 1 33
A Natural Cosmic Calendar
Ian Beardsley
Genesis Project California 2022!
⋅
⋅
of 2 33
Here we highlight some of the results we could consider the building blocks of creation;
the hydrocarbons which are the skeletal structure of life chemistry, carbon the core
element of life, the radius of the proton from which the elements are made, the radius of
the solar system and the relative abundances of the most abundant elements in the
universe hydrogen and helium from which all the others were made by stars in
nucleosynthesis. We present the final results, but the full calculations are done in the
appendices. The whole thing unfolds around a six-fold symmetry. In Appendix 1 we
suggest the existence of a Natural Cosmic Calendar.
Contents:
1.0 Overview………………………………………………..3
2.0 Creation…………………………………………………5
3.0 Geometric Transform…………………………………10
Appendix 1 Time Unit……………………………………..14
Appendix 2 Solar System Size…………………………..22
Appendix 3 Primordial Abundances……………………26
Appendix 4 Solar Eclipse.…….………………………….29
Appendix 5 Code For Program………………………….32!
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1.0 Overview
Equation 1.1 !
Where h is Planck’s constant 6.62607E—34 Js, is the radius of a proton 0.833E-15m, G is
the universal constant of gravitation 6.67408E-11 (Nm2)/(kg2), and c is the speed of light
299,792,459 m/s. And is t=1 second. is the Sommerfeld constant (or fine structure
constant) is 1/137. The mass of a proton is .!
Since plank’s constant h is a measure of energy over time where space and time are
concerned it must play a role. Of course the radius of a proton plays a role since squared and
multiplied by it is the surface area of our proton embedded in space. The gravitational
constant is force produced per kilogram over a distance, thus it is a measure of how the
surrounding space has an effect on the proton giving it inertia. The speed of light c has to play
a role because it is the velocity at which events are separated through time. The mass of a
proton has to play a role because it is a measurement of inertia itself. And alas the fine
structure constant describes the degree to which these factors have an effect. We see the
inertia then in equation 1.1 is six protons over 1 second, by dimensional analysis. !
The fine structure constant squared is the ratio of the potential energy of an electron in the first
circular orbit to the energy given by the mass of an electron in the Bohr model times the speed
of light squared: !
!
Matter is that which has inertia. This means it resists change in position with a force applied to
it. The more of it, the more it resists a force. We understand this from experience, but what is
matter that it has inertia?!
In this analogy we are suggesting a proton is a three dimensional bubble embedded in a two
dimensional plane. As such there has to be a normal vector holding the higher dimensional
sphere in a lower dimensional space. Thus if we apply a force to to the cross-section of the
sphere in the plane there should be a force countering it proportional to the normal holding it in
a lower dimensional universe. This counter force would be experienced as inertia.!
1
t
1
α
2
m
p
h4πr
2
p
Gc
= 6protons
r
p
t
1
α
m
p
= 1.67262E − 27kg
4π
α
2
=
U
e
m
e
c
2
of 4 33
#
of 5 33
2.0 Creation
Equation 2.1. !
Where t1 is one second, and six protons is carbon:!
!
And,…!
!
Is proton-seconds. Divide by time we have a number of protons because it is a mass divided
by the mass of a proton. But these masses can be considered to cancel and leave pure
number. We make a program that looks for close to whole number solutions so we can create a
table of values for problem solving (See Appendix 5 page 29).!
By what value would you like to increment?: 0.25!
How many values would you like to calculate for t in equation 1 (no more than 100?): 100!
24.1199 protons 0.250000 seconds 0.119904 decpart !
12.0600 protons 0.500000 seconds 0.059952 decpart !
8.0400 protons 0.750000 seconds 0.039968 decpart !
6.0300 protons 1.000000 seconds 0.029976 decpart !
4.0200 protons 1.500000 seconds 0.019984 decpart !
3.0150 protons 2.000000 seconds 0.014988 decpart !
2.1927 protons 2.750000 seconds 0.192718 decpart !
2.0100 protons 3.000000 seconds 0.009992 decpart !
1.2060 protons 5.000000 seconds 0.205995 decpart !
1.1486 protons 5.250000 seconds 0.148567 decpart !
1.0964 protons 5.500000 seconds 0.096359 decpart !
1.0487 protons 5.750000 seconds 0.048691 decpart !
1.0050 protons 6.000000 seconds 0.004996 decpart !
0.2487 protons 24.250000 seconds 0.248659 decpart !
0.2461 protons 24.500000 seconds 0.246121 decpart !
0.2436 protons 24.750000 seconds 0.243635 decpart !
A very interesting thing here is looking at the values generated by the program, the smallest
integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds
produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest
of the elements heavier than carbon are formed by fractional seconds. These are the
hydrocarbons the backbones of biological chemistry.!
1
t
1
α
2
m
p
h4πr
2
p
Gc
= 6protons
α
2
=
U
e
m
e
c
2
1
α
2
m
p
h4πr
2
p
Gc
of 6 33
Equation 2.2 A !
Equation 2.2 B !
Equation 2.3. !
Equation 2.4. !
Equation 2.5. !
Where the Lunar orbital period is 27.32 days but the lunar month in the denominator can be
that or as much as the 31 day month. is carbon (See Appendix 1 page 14).!
Equation 2.6. !
Equation 2.7. !
is the radius of a proton. k=18/3=6. We multiply by 1 second for carbon. Thus we see since
the periodic table is periodic over 18 groups it is deeply connected to the six that occurs in the
guzals of India, the sixes of the Middle Eastern and Persian music, and the six of Spain’s
flamenco. Indeed the key to playing in this meter is in understanding the relationship between 3
and 2 which defines the syncopated nature of the these styles, which is the reconciliation
between these two numbers. It occurs by way of the factor of six over a period of 18. Eighteen
as I said is the periodicity of the periodic table of the elements, and I would like to say now that
carbon (C) is the sixth element and the core element of life because it becomes when it
combines with other elements, which is for the skeleton of life the hydrocarbons by combing
with hydrogen , which is one proton. I have found that carbon (C) predicts the radius of this
proton.!
Thus we see the dynamics of these music forms are based on six in a way that reconciles 2
with 3. Eighteen is the number that reconciles 2 with 3, in the periodic table by way of the
factor of 6 (carbon) the core element of life. That is:!
1
α
2
m
p
h4πr
2
p
Gc
= (6protons)(1second )
1
α
2
m
p
h4πr
2
p
Gc
= (1proton)(6secon ds)
1
6α
2
m
p
h4πr
2
p
Gc
= 1.004996352secon d s
1secon d ≈
(K . E . Moon)(Lun arOrbitalPer iod )
LunarMonth
EarthD ay
(K . E . Ear th)
1
α
2
m
p
h4πr
2
p
Gc
≈ ℂ
(K . E . Moon)(LunarOrbitalPer iod )
LunarMonth
EarthD ay
(K . E . Ear th)
ℂ
3
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = 18
r
p
= kα
2
m
p
Gc
4πh
r
p
C
4−
H
+
of 7 33
3+3+3=9!
2+2+2=6!
!
!
!
The formulation of proton-seconds:!
!
!
Divide by time and we have a number of protons because it is a mass divided by the mass of a
proton. But these masses can be considered to cancel and leave pure number. We have that!
Equation 2.8 A !
Equation 2.8 B !
Interestingly 78% is the percent of N2 in the atmosphere and 21% is the percent of O2 in the
atmosphere (by volume). These are the primary constituents that make it up. The rest is
primarily argon and CO2. This gives the molar mass of air as a mixture is:!
Equation 2.9. !
Equation 2.10. !
Equation 2.11. !
is the radius of the solar system. Ge and Si in density g/cm2 not g/cm3 (thin disk). (See
Appendix 2 page 19)!
3 ⋅ 6 = 18
2 ⋅ 9 = 18
2 ⋅ 3 = 6
1
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
1
α
2
m
p
h4πr
2
p
Gc
6
∫
3
2
1
t
2
dt = 6
(
1
2
−
1
3
)
= 0.78
∫
3
2
cos
−1
(x /2)d x =
1
6
(
3π − 6
)
−
π − 4
2 2
= 0.21
0.78N 2 + 0.21O 2 = 29.0g/m ol
air
H
2
O
≈ Φ
R
s
=
3M
p
π(0.78Ge + 0.21Si)
R
s
of 8 33
We can produce the relative abundances of the primordial elements hydrogen and helium in
the universe (See Appendix 3 page 23).!
Where !
Or. !
!
!
Thus for hydrogen:!
Equation 2.12 A. =74%.!
For Helium:!
Equation 2.12. B =26%!
This is interesting because the Universe is about 74% Hydrogen and 24% Helium, the rest of
the elements making up the other 2%. Thus we can say or . We have
multiplied the first by 1 for Hydrogen element 1 (1 proton), and the second by 2 for helium
element 2 (2 protons). In a sense then, the probabilities represent the probability of finding
hydrogen and helium in the Universe. Hydrogen and much of the helium were made
theoretically in the Big Bang of the big bang theory. The other elements were synthesized from
these by the stars. The difference between hydrogen and helium is that some of the helium was
ψ
2
= ex p
[
−
x
2
d
2
⋅
1
1 + t
2
/τ
2
]
τ =
m d
2
ℏ
d =
Si − C
C
=
4
3
d
2
=
16
9
ψ
2
= ex p −
C
2
x
2
(Si − C )
2
⋅
1
1 +
[
ℏC
2
m(Si − C )
2
]
2
t
2
ψ
2
= ex p −
9
16
x
2
⋅
1
1 +
ℏ
2
81
m
2
256
t
2
ψ
2
= (1)ex p −
9
16
(1proton)
2
⋅
1
1 +
(0.075)81
(1)256
(6secon ds)
2
ψ
2
= (2)ex p −
9
16
(2proton)
2
⋅
1
1 +
(0.075)81
(4)256
(3secon ds)
2
ℏ
2
= 0.075
ℏ = 0.27386
of 9 33
made in stars, that may be why it is 26% not 24% because it could include the remaining 2%
of elements in the universe.!
We see we have the following theorem:!
Theorem 2.1. !
So as an example,…!
!
Is fluorine (F). Divide by xy with x=y=1 and we have current density. And multiply by 1 second
which is carbon and we have protons per square meter. Charge conservation gives:!
!
Is 9-9=0. In general…!
!
!
#
1
α
2
m
p
h4πr
2
p
Gc
∫
t
dt
t
3
=
∬
S
J ⋅ d
S
1
α
2
m
p
h4πr
2
p
Gc
∫
1.0
0.5
dt
t
3
=
∬
S
J ⋅ d
S = − 3
(
1 −
1
0.25
)
= 9
proton s
secon d
∂ρ
∂ t
+ ∇ ⋅
J = 0
J (x, y, z) = (0,0,J ) = − J
k
d
S = d x d y
k
J ⋅ d
S = (0,0,J ) ⋅ (0,0,d x d y) = − Jd x d y
of 10 33
3.0 Geometric Transform
The form we have is based on six in a way that reconciles 2 with 3. Eighteen is the number that
reconciles 2 with 3, in the periodic table by way of the factor of 6 (carbon) the core element of
life. The periodic table is cyclical over 18 groups. That is:!
!
!
is the radius of a proton. We multiply by 1 second for carbon. We have 1 second in earth-
moon-sun orbital mechanics:!
!
!
3+3+3=9!
2+2+2=6!
!
!
!
We can bring all of this together with one equation by considering the square and equilateral
triangle. If we draw in the diagonal of the square we make an angle of 45 degrees. This means
its hypotenuse is . The cosine of 45 degrees is . We have!
!
If we draw in the altitude of the the equilateral triangles formed by triangles made by the
equilateral triangle’s radii, we have its radius is . This means!
!
3
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = 18
r
p
=
18
3
α
2
m
p
Gc
4πh
(1secon d )
r
p
1secon d ≈
(K . E . Moon)(LunarOrbitalPer iod )
LunarMonth
EarthD ay
(K . E . Ear th)
1
α
2
m
p
h4πr
2
p
Gc
≈ ℂ
(K . E . Moon)(LunarOrbitalPer iod )
LunarMonth
EarthD ay
(K . E . Ear th)
3 ⋅ 6 = 18
2 ⋅ 9 = 18
2 ⋅ 3 = 6
2
2/2
2 cos
π
4
1/ 3
2 cos
π
6
of 11 33
Theorem 4.1. If we add the half-square to the square it is the counterpart to the half of the
equilateral triangle making a rectangle (See illustration next page). This half triangle if we draw
in the circle with the radius of its shorter side produces the golden ratio . The golden ratio is:!
!
This results in the following unifying equation:!
!
n=4,5,6!
We can move between these forms geometrically (Following illustration).!
!
Φ
2 cos
π
5
f (n) = 2cos
π
n
of 12 33
In that the constant predicts the radius of a proton with the unit of a second we have to ask
why the second has the duration it has. The second comes to us from the Ancient Greeks
dividing the degree of a circle into 60 minutes and that into 60 seconds because they used a
sexagesimal (base 60) counting system, and this might be the answer because 60 was
probably chosen because it is evenly divisible by 1,2,3,4,5,6..12, 15, 20, 30… and the Ancient
Greeks got that from the Babylonians who got it from the Sumerians. Consider then that the
second comes from 365 days in a year, 12 hours a day, 60 minutes in an hour, 60 seconds in a
minute. Thus the calendar formed like this (sexagesimal) could be related to the nature of solar
system formation because the 12 moons per year (earth orbital period) divided up by
sexagesimal reconcile these periods (See Appendix 4 Page 26).!
#
of 13 33
Since we have assigned time values to numbers of protons, and thus elements, from equation
1.1:!
!
And written a program that determines them (See Appendix 5 page 29) :!
24.1199 protons 0.250000 seconds 0.119904 decpart !
12.0600 protons 0.500000 seconds 0.059952 decpart !
8.0400 protons 0.750000 seconds 0.039968 decpart !
6.0300 protons 1.000000 seconds 0.029976 decpart !
4.0200 protons 1.500000 seconds 0.019984 decpart !
3.0150 protons 2.000000 seconds 0.014988 decpart !
2.1927 protons 2.750000 seconds 0.192718 decpart !
2.0100 protons 3.000000 seconds 0.009992 decpart !
1.2060 protons 5.000000 seconds 0.205995 decpart !
1.1486 protons 5.250000 seconds 0.148567 decpart !
1.0964 protons 5.500000 seconds 0.096359 decpart !
1.0487 protons 5.750000 seconds 0.048691 decpart !
1.0050 protons 6.000000 seconds 0.004996 decpart !
0.2487 protons 24.250000 seconds 0.248659 decpart !
0.2461 protons 24.500000 seconds 0.246121 decpart !
0.2436 protons 24.750000 seconds 0.243635 decpart !
And we have angles associated with the elements from their geometry, we have a the equation
of a wave:!
Equation 3.1. !
If we say is the amplitude and t=1 second is carbon, then trying
degrees from our triangles and squares:!
!
!
!
And for hydrogen with a value of 6 seconds, is:!
!
!
!
This is interesting to me because hydrogen being the basis of the periodic table is what it
should be, oscillating between -1 and 1 with the triangle and is 0 for the square. Carbon being
the 6 factor crux of the periodic table and core element of life chemistry is the dynamic ratios
involving the square root of three, square root of two, and one half.#
1
t
1
α
2
m
p
h4πr
2
p
Gc
= 6protons
A = A
0
cosωt
A
0
= 1
ω = 30,60,45
A(60) = cos(60 ⋅ 1) = 0.5
A(30) = cos(30 ⋅ 1) = 3/2
A(45) = cos(45 ⋅ 1) = 2/2
A(60) = cos(60 ⋅ 6) = 1
A(30) = cos(30 ⋅ 6) = − 1
A(45) = cos(45 ⋅ 6) = 0
of 14 33
Appendix 1
Buckminster Fuller writes in Synergetics:!
205.01 The geometrical model of energy configurations in synergetics is developed from a
symmetrical cluster of spheres, in which each sphere is a model of a field of energy all of whose
forces tend to coordinate themselves shuntingly or pulsatively, and only momentarily in
positive or negative asymmetrical patterns relative to, but never congruent with, the eternality of
the vector equilibrium…This closest packing of spheres in 60-degree angular relationships
demonstrates a finite system in Universal geometry.
of 15 33
#
of 16 33
Thus we see carbon is 8 tetrahedra and 6 octahedra, each of which is 2 tetrahedra I found
them precisely this in their two dimensional analogs the regular dodecahedron and regular
octagon:!
We see that the atomic radius of silicon the core element of artificial intelligence (transistor
technology) fits together with the core element of biological life carbon if the silicon is taken as
inscribed in a regular dodecagon and the carbon is taken as inscribed in a regular octagon. We
have:!
, , !
, , !
Apothem: !
For a regular dodecagon:!
!
The radius of a silicon atom is Si=0.118nm and that of carbon is
C=0.077nm:!
!
!
!
This has an accuracy %!
But, we still have the question of 1 second in equation 2.1:!
!
We see that the vector equilibrium employs 60-degree coordination. And that it involves a layer
of 12 spheres. However we have a 12 hour daytime and and 12 hour night time, which is a 24
hour day (Earth rotates through 15 deg an hour, dodecagon). But there are 60 minutes in an
hour and 60-degree coordination is connected to that in that the earth rotates through 360
degrees in 24 hours and 360/60 degrees is our six-fold symmetry of the periodic table. Each
hour then of the 24 hours is divided by 60 minutes and each minute divided by 60 is 1 second.
D = 1 + 2x
x
2
+ x
2
= 1
2
2x
2
= 1
2x
2
= 1
x = 2/2
D = 1 + 2
a = (1 + 2)/2 = 1.2071
a =
s/2
ta n(θ /2)
=
0.5
ta n(15
∘
)
= 1.866
Si
C
=
0.118
0.077
= 1.532
a
S
i
a
C
=
1.866
1.2071
= 1.54585
1.532
1.54585
100 = 99
1
t
1
α
2
m
p
h4πr
2
p
Gc
= 6protons
of 17 33
We have our one second. This layout of how we measure time would suggest that the unit of
one second is a natural division, and that it should therefore be a product of the orbital
mechanics of Earth-Moon-Sun system. The second, which came from reconciling the lunar
month with the 365 day earth orbital period. It is clear how to approach this.!
We need to consider the translational kinetic energy of the Earth. 1AU=1.496E11m and
. 1 year=3.1558E7s and,
. m=5.972E24kg. Here we would use:!
!
We find the same for the moon. The moon is 3.85E8m on average from the Earth center.
. Its orbital period T, is . We have
. We have the translational kinetic energies of the Earth and
moon are:!
Earth: !
Moon: !
I found we don’t have to add on the rotational kinetic energies to the translational kinetic
energies because they are too small to have an effect within the accuracies we are working.
What we do now is create a sort of Planck’s constant h for the moon by multiplying its kinetic
energy over the time for a complete orbit around the earth:!
!
And we divide this by the Earth’s kinetic energy to get:!
!
This is the interesting thing here, we don’t need to include the rotational kinetic energies and
and one second is obtained by dividing our result by . That is:!
!
and !
This makes sense. Because we need to bring in the lunar orbital period of 27.32 days, which
defines the month for which there are twelve in a year as compared to the Earth’s rotation
which is once per day. We have!
!
2π (1.496E11m) = 9.3996E11m eters /year
v = 2.9785E4meters /secon d
E =
1
2
mv
2
2π (3.85E8m) = 2.419E 9m
T = 27.32d ays = 2.36E6s
v = 1.025E 3m /s ≈ 1000m /s
E = 2.649E33Joules
E = 3.67E 28Joules
h = (3.67E 28J )(2.36E6s) = 8.6612E 34J ⋅ s
8.6612E 34J ⋅ s
2.649E33J
= 32.696secon ds
π
3
32.696secon ds
π
3
= 1.05s ≈ 1secon d
π
3
= 31.006
1
π
3
= 0.03225
27.32d ays
1day
= 27.32
of 18 33
This is close to . But the calendar month is 30 days, except for February, which is 28 days,
and 29 every four years for leap year, and January, March, May, July, and August are 31 days.
is almost exactly the 31 day month that occurs 5 months a year. The leap year is every four
years because the Earth year is not exactly 365 days but is 365.25 days.!
Let us now call upon equation 2.1:!
!
And write it!
!
We have:!
!
And we finally write:!
!
Where 6 protons is carbon the basis of life. We might write it:!
!
In order for this last equation to work perfectly the earth day has to be shorter. And indeed a
long time ago it was. But how long ago? Our Equation becomes:!
!
(24hours)/(1.2)=20 hours/day!
Over the past several billion years the length of the year has not deviated much from 365.25
days because of Kepler’s period for the orbit of a planet but, because the earth loses energy to
the moon, its days becomes longer over time by 0.0067 hours per million years which, we can
π
3
π
3
1
t
1
α
2
m
p
h4πr
2
p
Gc
= 6protons
1
α
2
m
p
h4πr
2
p
Gc
= (6protons)(1secon d )
1secon d ≈
(K . E . Moon)(LunarOrbitalPer iod )
LunarMonth
EarthD ay
(K . E . Ear th)
1
α
2
m
p
h4πr
2
p
Gc
≈ (6protons)
(K . E . Moon)(LunarOrbitalPer iod )
LunarMonth
EarthD ay
(K . E . Ear th)
1
α
2
m
p
h4πr
2
p
Gc
≈ ℂ
(K . E . Moon)(LunarOrbitalPer iod )
LunarMonth
EarthD ay
(K . E . Ear th)
1
α
2
m
p
h4πr
2
p
Gc
≈ (6protons)
(K . E . Moon)(Ear th Da y)
(K . E . Ear th)
= 1.2seconds
of 19 33
see from examining sedimentation band growth, which follows the lunar month. To get our 20
hours per day we have to go back about 600 million years:!
24-20=0.0067t!
t=597 million years!
What was going on then? It turns out it was a very important time, when the Earth went
through a drastic change which lead to an explosion of life, and we started to see more
complex life.!
Let’s talk about what is going on here:!
The reason is there. I am showing the unit of a second comes from the orbital mechanics of
the Earth-Moon-Sun System, the orbital period of the moon to the rotation period of the earth
is an approximation to pi cubed, The lunar month is determined by the orbital period of the
moon. The lunar month can be 32 days. This divided by the one day rotation of the earth is 32
is approximately pi cubed. In the equation when the lunar orbital period equals the lunar month,
they cancel and leave the kinetic energy of the moon divided by the kinetic energy of the earth
times the earth day. For it to work perfectly it means that equals one second, but it is 1.2
seconds so the earth day needs to be shorter, which it was by the right amount about 600
million years ago. That is how I got that time in earth history, which seems to be when the earth
went through a drastic change leading to complex life, according to the geologic record. Which
works well for the equation because this is a beginning time for complex life, and it seems to be
the beginning time for the synced calendar in terms of the equation. We don’t need to put pi
there because we have what we need to explain one second in terms of the Earth-Moon-Sun
Orbital mechanics. Google writes for the time 600 million years ago:
“A global ice age over 600 million years ago dramatically altered the face of the planet, leaving a
barren, flooded landscape and clear oceans, according to a study that may have important
implications for the evolution of complex life."
“The ancient sea of 600 million years ago was not soupy, the researchers said. Instead, it had a
consistency and nutrient level similar to today's oceans. This would have had major implications
for the story of animal evolution.”
We can take this further.!
In section 3.0 page 13 we had equation 3.1 which was:!
Equation 3.1. !
Which with we write!
!
In Appendix 1 page 18 we had!
π
A = A
0
cosωt
A
0
= 1
A = cosωt
of 20 33
!
We needed it to be one second so the earth day had to be shorter. We went back in time to the
point where the earth day gave 1 second because 1 second is carbon giving the radius of a
proton. I would suggest this time to be the starting point of a cosmic calendar. It turned out to
be 597 million years ago an age which is perfect for such an idea because this was the
Cambrian age of the famed Cambrian explosion where the earth underwent some kind of
drastic change leading to an explosion of complex life. With carbon the core element of
biological chemistry and our equation equal to one second that gives the radius of the proton it
couldn’t work better. !
But, we ask what other age was pivotal to human development. It would be the end of the
Cretaceous Period when the dinosaurs went extinct 65 million years ago due to an asteroid
hitting the earth in Yucatan. I say this because we know the extinction of the dinosaurs gave
small mammals the opportunity to evolve into humans. To find out how long the earth day was
65 million years ago we write:!
24-x=0.0067t!
And find!
x=23.5645 hours!
For the length of an earth day. That is it was less by1-0.5645=0.4355 of an hour. This is close
to . We have already established that the zero of our calendar
was when when the earth day was 20 hours during the Cambrian explosion. 21 hours for an
epoch would give:!
24-21=0.0067t with t=447.76 million years is approximately 450 million years!
24-22=0.0067t with t=298.5 million year is approximately 300 million years!
24-23=0.0067t with t=149.25 million years is approximately 150 million years.!
24-23=1 and is . We have!
!
We say 20 hours+3 hours is 0 hours+3 hours since 20 hours is the zero. We have!
=dinosaur extinction!
= !
The 0.5645 hours beyond 23 hours is , . Where
is the radius of an equilateral triangle to its side. Thus we see a correlation between the end of
the dinosaurs and the beginning of mammalian life in our calendar zeroed with the Cambrian
1
α
2
m
p
h4πr
2
p
Gc
≈ 6
(K . E . Moon)
(K . E . Ear th)
(Ear th Day) = 1.2secon ds
(1/2)cos30
∘
= 0.4330 = 3/4
cosθ = 1
θ = 0
∘
cos(0
∘
) −
2
3
cos(30
∘
) =
1
2
cos(30
∘
)
3cos(0
∘
) +
2
3
cos(30
∘
)
3hours +
3
3
hours
cosθ = 0.5645 ≈ 1/ 3
θ = 54.7356
∘
1/ 3
of 21 33
Explosion. I am beginning to think there is some kind of a Natural Cosmic Calendar. If we
guess at what the pattern is here we might guess after the dinosaurs went extinct and
mammals could evolve towards humans, then the next term in our equation might be . In
which case we have:!
!
=20+3+0.57735+0.4714=24.04875 hours!
Is exactly the length of the Earth day in present times to one place after the decimal.!
Can we go back in our cosmic calendar to periods before the Cambrian explosion. 65 million
years ago of course is not a precise date for the very day on which an asteroid hit earth
bringing about the extinction of dinosaurs, but given the vast expanse of time we are working
with percentage-wise that estimate is very good for our purposes. Our calendar starts at the
Cambrian explosion. What major event existed before that? It would seem about 2 billion years
ago plants and plankton came into extraordinary abundance, and these photosynthesizers
converted the primordial CO2 in the atmosphere, making it into oxygen, which they continue to
do to this day. As a result, about two billion years ago the Earth atmosphere was converted to
one having a high percentage of oxygen gas (O2). Thus a new kind of organism came into
existence, one that gets its energy from food by burning it in oxygen.!
Today many organisms use copper (Cu2++) for its metabolism because it is much more soluble
than iron in its reduced state. Iron is quickly oxidized to its ferric state and in this state is
extremely insoluble. Thus, before the oxygen rich atmosphere, organisms could use iron.!
The interesting thing is we don’t have to look just at the paleontological record to ascertain the
history of life life on Earth, we can find the metabolic pathways common to all life today and we
have found in common all life has certain things. We have found one of those things is an iron-
based metabolism. Thus we have a way of finding when the common ancestor to all life
existed. We call her LUCA. Since she had an Iron based metabolism she existed at least before
the explosion of photosynthesizes, which means she is at least 2 billion years old. But we have
a fossil record that goes back 3.5 billion years for the first life, and everyday that keeps getting
pushed back, even to 4 billion years, or more. The Earth and Sun formed about 5 billion years
ago.!
So we may not have the data to take our calendar back to a date for LUCA. But perhaps we
can call 2 billion years accurate for the explosion of photosynthesizers and the arrival of an
oxygen-rich atmosphere.!
2/3
20hours + 3h ours +
3
3
hours +
2
3
hours
of 22 33
Appendix 2
Indeed I find the size of the solar system is connected to Silicon through its counterpart as a
semiconductor material germanium. Let’s look at that:!
While we have!
!
Is a number of protons!
!
Is proton-seconds. Divide by time we have a number of protons because it is a mass divided
by the mass of a proton. But these masses can be considered to cancel and leave pure
number. We have that!
!
!
Interestingly 78% is the percent of N2 in the atmosphere and 21% is the percent of O2 in the
atmosphere (by volume). These are the primary constituents that make it up. The rest is
primarily argon and CO2. This gives the molar mass of air as a mixture is:!
!
Now interestingly, I have found this connected to the solar system by considering a mixture of
silicon, the primary constituent of the Earth crust, and germanium just below it in the periodic
table, in the same proportions of 78% and 21%. Silicon is also the primary second generation
semiconductor material (what we use today) and germanium is the primary first generation
semiconductor material (what we used first). The silicon is directly below our carbon of one
proton-second, silicon directly below that, and germanium directly below that, in the periodic
table. So we are moving directly down the periodic table in group 14. The density of silicon is
2.33 g/cm3 and that of germanium is 5.323 g/cm3. Let us weight these densities with our 0.21
and 0.78:!
!
Now consider this the starting point for density of a thin disc decreasing linearly from the Sun
to Pluto (49.5AU=7.4E14cm). Thus,…!
1
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
1
α
2
m
p
h4πr
2
p
Gc
6
∫
3
2
1
t
2
dt = 6
(
1
2
−
1
3
)
= 0.78
∫
3
2
cos
−1
(x /2)d x =
1
6
(
3π − 6
)
−
π − 4
2 2
= 0.21
0.78N 2 + 0.21O 2 = 29.0g/m ol
0.21Si + 0.78Ge = 4.64124g/cm
3
of 23 33
!
Thus,…!
!
Or,…!
!
Thus,…!
!
If we add up the masses of the planets in our solar system they are 2.668E30 grams.!
Since!
!
Meaning mixing germanium and silicon in the same proportion that occurs with N2 and O2 in
the atmosphere and with!
!
!
Where!
!
In the first integral. See the following pages to see the computation of the mass of the planets
in the solar system…!
!
ρ(r) = ρ
0
(
1 −
r
R
)
M =
∫
2π
0
∫
R
0
ρ
0
(
1 −
r
R
)
rdrdθ
M =
πρ
0
R
2
3
M =
π(4.64124)(7.4E14)
2
3
= 2.661E 30
2.661
2.668
(100) = 99.736
6
∫
3
2
1
t
2
dt = 6
(
1
2
−
1
3
)
= 0.78
∫
3
2
cos
−1
(x /2)d x =
1
6
(
3π − 6
)
−
π − 4
2 2
= 0.21
6 =
1
α
2
m
p
h4πr
2
p
Gc
of 24 33
As we can see Jupiter carries the majority of the mass, Saturn a pretty large piece, and
somewhat large Uranus and Neptune. We don’t even list Pluto’s mass. When we consider!
!
!
We are considering and . These come from!
!
!
From 30 degrees to 45 degrees is 15 degrees. The Earth rotates through 360/24 is 15 degrees
per hour. The hour is divided into 60 minutes and minutes into 60 seconds…!
We have said!
!
6
∫
3
2
1
t
2
dt = 6
(
1
2
−
1
3
)
= 0.78
∫
3
2
cos
−1
(x /2)d x =
1
6
(
3π − 6
)
−
π − 4
2 2
= 0.21
2
3
2 cos(45
∘
) = 2
2 cos(30
∘
) = 3
M =
πρ
0
R
2
3
of 25 33
For a thin disc. Thus we have a definition for the radius of the solar system, :!
!
Where!
!
!
!
!
is the mass of all the planets. We also have that the molar mass of air to the molar mass of
water is approximately the golden ratio. The interesting thing is we determine a definition for
the radius of the solar system and predict the hydrocarbons (backbones of life) all in one fell
swoop. Thus since if the universe is mostly hydrogen and helium if we can predict their relative
abundances with proton-seconds then we can feel quite certain we are formulating things
correctly, consider the following gaussian distribution illustrated…#
R
s
R
s
=
3M
p
π(0.78Ge + 0.21Si)
1
α
2
m
p
h4πr
2
p
Gc
∫
3
2
1
t
2
dt = 0.78
∫
3
2
cos
−1
(x /2)d x = 0.21
air = 0.78N
2
+ 0.21O
2
air
H
2
O
≈ Φ
M
p
of 26 33
Appendix 3
We consider a Gaussian wave-packet at t=0:!
!
We say that d which in quantum mechanics would be the delocalization length when squared is
. A is the amplitude and we might say it is . We write the wave packet as a
Fourier transform which is:!
!
We use the identity that gives the integral of a quadratic:!
!
Solve the equation!
!
With the initial condition!
!
A plane wave is the solution:!
!
Where, !
The wave-packet evolves with time as!
!
Calculate the Gaussian integral of dp!
and !
ψ (x,0) = Ae
−
x
2
2d
2
Si − C
C
Ga − P
C
ψ (x,0) = Ae
−
x
2
2d
2
=
∫
dp
2π ℏ
ϕ
p
e
i
h
px
∫
∞
−∞
e
−α
2
x+βx
d x =
π
α
e
β
2
4α
iℏ∂
t
ψ (x, t) =
p
2m
ψ (x, t)
ψ (x,0) =
∫
dp ⋅ e
p
2
d
2
2ℏ
2
⋅ e
−
i
h
px
e
i
ℏ
( px−ϵ( p)t)
ϵ( p) =
p
2
2m
ψ (x, t) =
∫
dp ⋅ e
p
2
d
2
2ℏ
2
⋅ e
−
i
h
( px−
p
2
2m
t)
α =
d
2
2ℏ
2
+
it
2mℏ
β =
i x
ℏ
of 27 33
The solution is:!
where !
We notice here one of the things you can do with our equation for proton-seconds is integrate
from 0.5 sec to 1 sec and you get one which multiplied by the constant which is six yields six.
Now look up 0.5 seconds from the data output from the program and it is magnesium, then go
to one second and it is carbon, thus the integral from magnesium in time to carbon in time is
carbon in protons. Now consider life as we know is based on carbon because it has four
valence electrons, but it is not based on silicon, which has four valence electrons as well,
because in the presence of oxygen it readily forms SiO2 (sand or glass) leaving it unavailable to
nitrogen, phosphorus, and hydrogen to make make amino acids the building blocks of life. But
silicon can be doped with phosphorus, boron, gallium, and arsenic to make semiconductors --
transistor technology from which we can build artificial life (artificial intelligence, AI). We can
integrate over many time ranges to explore millions of more facets to the equation:!
!
!
!
We should say!
or !
The way I am using equation 1 is . We have:!
Equation 3.14!
!
ψ
2
= ex p
[
−
x
2
d
2
⋅
1
1 + t
2
/τ
2
]
τ =
m d
2
ℏ
1
α
2
m
p
h4πr
2
p
Gc
∫
t
C
t
Mg
1
t
2
dt = 6 = carbon(C )
t
Mg
= 0.5secon ds
t
C
= 1secon d
d =
Si − C
C
=
4
3
d
2
=
16
9
τ = d
2
ψ
2
= ex p −
C
2
x
2
(Si − C )
2
⋅
1
1 +
[
ℏC
2
m(Si − C )
2
]
2
t
2
of 28 33
Thus for hydrogen:!
!
=74%.!
For Helium:!
=26%!
This is interesting because the Universe is about 74% Hydrogen and 24% Helium, the rest of
the elements making up the other 2%. Thus we can say or . We have
multiplied the first by 1 for Hydrogen element 1 (1 proton), and the second by 2 for helium
element 2 (2 protons). In a sense then, the probabilities represent the probability of finding
hydrogen and helium in the Universe. Hydrogen and much of the helium were made
theoretically in the Big Bang of the big bang theory. The other elements were synthesized from
these by the stars. The difference between hydrogen and helium is that some of the helium was
made in stars, that may be why it is 26% not 24% because it could include the remaining 2%
of elements in the universe. It must be kept in mind the data we have on the universe as a
whole for relative abundances can only currently be ballpark figures.#
ψ
2
= ex p −
9
16
x
2
⋅
1
1 +
ℏ
2
81
m
2
256
t
2
ψ
2
= (1)ex p −
9
16
(1proton)
2
⋅
1
1 +
(0.075)81
(1)256
(6secon ds)
2
ψ
2
= (2)ex p −
9
16
(2proton)
2
⋅
1
1 +
(0.075)81
(4)256
(3secon ds)
2
ℏ
2
= 0.075
ℏ = 0.27386
of 29 33
Appendix 4
Kepler’s Law of planetary motion is !
Equation 1. !
For the Sun with T the orbital period of the planet and its distance from the Sun in
astronomical units, for circular orbits. For other stars we have to include a constant k involving
the masses of the bodies:!
Equation 2. !
If the mass of the body orbiting the Star m is is small compared to the mass of the star it is
orbiting we have!
Equation 3. !
The Earth is the third planet, is in the habitable zone, and its distance from the Sun defines 1
AU. Thus we ask: What is k for other star systems? For stars on the main sequence their
luminosity is proportional to their masses raised to the power of 3.5, as an estimate. We have:!
Equation 4: !
Where L is in solar luminosities and M is in solar masses:!
!
!
Further if we say since the Earth is in the right zone to be habitable ( ) then if a star is 100
times brighter than the Sun by the inverse square law its habitable zone is is
10 times further from the star it orbits than the Earth is from the Sun. We have!
Equation 5: !
Combining equations 4 and 5:!
T
2
= a
3
a
1
k
= G
M + m
4π
2
≈ G
M
4π
2
a
3
=
GM
4π
2
T
2
L =
(
M
M
⊙
)
3.5
L
⊙
= 3.9E 26J/s
M
⊙
= 1.98847E 30kg
ℍ
100 = 10AU
ℍ =
(
L
L
⊙
)
of 30 33
Equation 6: !
For another star system we can write equation 3!
Equation 7. !
Where n is the number of solar masses of the star and k has M in solar masses. Combining this
with equation 6 we have for the habitable zone of a star:!
Equation 8. !
Thus if the star is 100 times more luminous than the sun!
!
Or,…!
!
Yielding from equation 6!
!
And!
Equation 9. !
Which is 31.623 years for 3.72759. This result is (31.623 yr)(365.25 days)=11,550 days for the
orbital period. 10 AU from this star is about the distance Saturn is from the Sun (9.54AU), with
year equal to 10,759 days. The orbital period is about the same but will be shorter by the right
amount if we include because we have said the star is more massive so it is orbiting
faster to keep from falling in. Turns out to be a 16.38 year orbit.!
If in the planet that has life orbiting a star has an indication to its intelligence that there is a
mystery before it, just as we do on the Earth in that we have a moon that perfectly eclipses the
Sun as seen from the Earth, then…!
This is because:!
ℍ =
(
M
M
⊙
)
3.5
T
2
a
3
=
k
n
T
2
=
k
n
(
M
M
⊙
)
21
4
ln(100) = 3.5ln(M /M
⊙
)
M
M
⊙
= 3.72759
ℍ = 3.72759
3.5
= 10AU
T ∝
(
M
M
⊙
)
21/8
k /n
of 31 33
!
!
Which are approximately equal. As well we can look at it as:!
!
!
Which are about the same as well. The interesting thing is that since our ratios are around
0.0025 and 0.0045, then…!
!
I say this is interesting because this the ratio of the precious metal gold (Au) to that of silver
(Ag) by molar mass these elements being used for religious and ceremonial jewelry:!
!
We have:!
!
!
Taking the average between these:!
Equation 10. !
Where, is the lunar orbit, is the solar radius, is the earth orbit, and is the radius of
the moon. What this means is that the moon perfectly eclipses the sun because the solar
radius is 1.8 times greater than the lunar orbital radius. And interestingly gold is yellow, silver is
silver and the sun is gold, and the moon is silver, the moon perfectly eclipses the sun allowing
us to study its outer atmosphere.#
(lun ar − orbit)
(ear th − orbit)
=
384,400k m
149,592,870k m
= 0.00257
(lun ar − ra dius)
solar − ra dius
=
1,738.1
696,00
= 0.0025
(lun ar − ra dius)
(lun ar − orbit)
=
(1,738.1)
(384,400)
= 0.00452
solar − ra dius
ear th − orbit
=
696,000
149,597,870
= 0.00465
0.0045
0.0025
=
9
5
= 1.8
Au
Ag
=
196.97
107.87
= 1.8
(lun ar − ra dius)(ear th − orbit)
(lun ar − orbit)
2
= 1.759577590
(solar − ra dius)
2
(ear th − orbit)(lunar − ra dius
= 1.863
1
2
⋅
(
r
2
m
⋅ R
2
⊙
+ r
2
e
R
2
m
r
2
m
⋅ r
e
⋅ R
m
)
=
Au
Ag
r
m
R
⊙
r
e
R
m
of 32 33
Appendix 5
Here is the code for the program:!
#include <stdio.h>!
#include <math.h>!
int main(int argc, const char * argv[]) {!
!
int n;!
float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792459,protons[100],r=0.833E-15;!
!
do!
{!
printf("By what value would you like to increment?: ");!
scanf("%f", &increment);!
printf("How many values would you like to calculate for t in equation 1 (no more than 100?):
");!
scanf("%i", &n);!
}!
while (n>=101);!
{!
for (int i=0; i<n;i++)!
{!
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));!
!
int intpart=(int)protons[i];!
float decpart=protons[i]-intpart;!
t=t+increment;!
if (decpart<0.25)!
{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);!
}}}}!
of 33 33
The Author!
